Find the product:
(x –2y – 3)(x2 + 4y2 + 2xy – 3x + 6y + 9)
We have,
a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3
We know that,
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)]
Also,
If, (x + y + z) = 0
Then (x3 + y3 + z3) = 3xyz
Using this, we get
= [x + (-2y) + 3] [(x)2 + (-2y)2 + (3) – (x) (-2y) – (-2y) (3) – (3) (x)]
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
= a# + b3 + c3 – 3abc
Where,
x = a, b = -2y and c = 3
(x – 2y + 3) (x2 + 4y2 + 2xy – 3x + 6y + 9)
= (x)3 + (-2y)3 + (3)3 – 3 (x) (-2y) (3)
= x3 – 8y3 + 27 + 18xy